Question

How do you find the limit of `ln(lnt)` as `t->oo`?

Answer 1

The answer is `infty`

Explanation:

Since `ln(t)->infty` as `t->infty`, it follows that `ln(ln(t))->infty` as `t->infty` (though very, very, slowly).

To illustrate how slowly `ln(ln(t))->infty` as `t->infty`, you might ask: how big should `t` be so that `ln(ln(t))>10`? (for example)

To answer this, solve the inequality `ln(ln(t))>10` by exponentiation: `ln(ln(t))>10 <=> ln(t)>e^(10) <=> t>e^(e^(10)) approx 9.4 times 10^(9565)`

In general, in order for `ln(ln(t))` to be bigger than an arbitrary `M>0`, you need to choose `t` so large that `t>e^(e^(M))`.